Prove that:
(i) cos 55° + cos 65° + cos 175° = 0
(ii) sin 50° − sin 70° + sin 10° = 0
(iii) cos 80° + cos 40° − cos 20° = 0
(iv) cos 20° + cos 100° + cos 140° = 0
(v) $\sin \frac{5 \pi}{18}-\cos \frac{4 \pi}{9}=\sqrt{3} \sin \frac{\pi}{9}$
(vi) $\cos \frac{\pi}{12}-\sin \frac{\pi}{12}=\frac{1}{\sqrt{2}}$
(vii) sin 80° − cos 70° = cos 50°
(viii) sin 51° + cos 81° = cos 21°
(i) Consider LHS :
$\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}$
$=2 \cos \left(\frac{55^{\circ}+65^{\circ}}{2}\right) \cos \left(\frac{55^{\circ}-65^{\circ}}{2}\right)+\cos 175^{\circ} \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$
$=2 \cos 60^{\circ} \cos \left(-5^{\circ}\right)+\cos 175^{\circ}$
$=2 \times \frac{1}{2} \cos 5^{\circ}+\cos 175^{\circ}$
$=\cos 5^{\circ}+\cos 175^{\circ}$
$=2 \cos \left(\frac{5^{\circ}+175^{\circ}}{2}\right) \cos \left(\frac{5^{\circ}-175^{\circ}}{2}\right)$
$=2 \cos 90^{\circ} \cos 85^{\circ}$
$=0$
Hence, LHS = RHS.
(ii) Consider LHS :
$\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$
$=2 \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right)+\sin 10^{\circ} \quad\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) c \cos \left(\frac{A+B}{2}\right)\right\}$
$=2 \sin \left(-10^{\circ}\right) \cos 60^{\circ}+\sin 10^{\circ}$
$=2 \times \frac{1}{2} \sin \left(-10^{\circ}\right)+\sin 10^{\circ}$
$=-\sin 10^{\circ}+\sin 10^{\circ}$
$=0$
Hence, LHS = RHS.
(iii) Consider LHS :
$\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}$
$=2 \cos \left(\frac{80^{\circ}+40^{\circ}}{2}\right) \cos \left(\frac{80^{\circ}-40^{\circ}}{2}\right)-\cos 20^{\circ} \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$
$=2 \cos 60^{\circ} \cos 20^{\circ}-\cos 20^{\circ}$
$=2 \times \frac{1}{2} \cos 20^{\circ}-\cos 20^{\circ}$
$=\cos 20^{\circ}-\cos 20^{\circ}$
$=0$
Hence, LHS = RHS.
(iv) Consider LHS :
$\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}$
$=2 \cos \left(\frac{20^{\circ}+100^{\circ}}{2}\right) \cos \left(\frac{20^{\circ}-100^{\circ}}{2}\right)+\cos 140^{\circ} \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$
$=2 \cos 60^{\circ} \cos \left(-40^{\circ}\right)+\cos 140^{\circ}$
$=2 \times \frac{1}{2} \cos 40^{\circ}+\cos 140^{\circ}$
$=\cos 40^{\circ}+\cos 140^{\circ}$
$=2 \cos \left(\frac{40^{\circ}+140^{\circ}}{2}\right) \cos \left(\frac{40^{\circ}-140^{\circ}}{2}\right)$
$=2 \cos 90^{\circ} \cos 50^{\circ}$
$=0$
Hence, LHS = RHS.
(v) Consider LHS :
$\mathrm{LHS}=\sin \left(\frac{5 \pi}{18}\right)-\cos \frac{4 \pi}{9}$
$=\sin \left(\frac{5 \pi}{18}\right)-\cos \left(\frac{\pi}{2}-\frac{\pi}{18}\right)$
$=\sin \left(\frac{5 \pi}{18}\right)-\sin \left(\frac{\pi}{18}\right)$
$=2 \sin \left(\frac{\frac{5 \pi}{18}-\frac{\pi}{18}}{2}\right) \cos \left(\frac{\frac{5 \pi}{18}+\frac{\pi}{18}}{2}\right) \quad\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$
$=2 \sin \left(\frac{\pi}{9}\right) \cos \frac{\pi}{6}$
$=2 \sin \left(\frac{\pi}{9}\right) \cos \frac{\pi}{6}$
$=2 \times \frac{\sqrt{3}}{2} \sin \left(\frac{\pi}{9}\right)$
$=\sqrt{3} \sin \left(\frac{\pi}{9}\right)=\mathrm{RHS}$
Hence, LHS = RHS.
(vi) Consider LHS :
LHS $=\cos \frac{\pi}{12}-\sin \frac{\pi}{12}$
$=\cos \left(\frac{\pi}{2}-\frac{5 \pi}{12}\right)-\sin \frac{\pi}{12}$
$=\sin \left(\frac{5 \pi}{12}\right)-\sin \frac{\pi}{12}$$=\sin \left(\frac{5 \pi}{12}\right)-\sin \frac{\pi}{12}$
$=2 \sin \left(\frac{\frac{5 \pi}{12}-\frac{\pi}{12}}{2}\right) \cos \left(\frac{\frac{5 \pi}{12}+\frac{\pi}{12}}{2}\right) \quad\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$
$=2 \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right)$
$=2 \times \frac{1}{2} \times \frac{1}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}}$
Hence, LHS = RHS.
(vii) Consider LHS :
$s \operatorname{in} 80^{\circ}-\cos 70^{\circ}$
$=\sin 80^{\circ}-\cos \left(90^{\circ}-20^{\circ}\right)$
$=\sin 80^{\circ}-\sin 20^{\circ}$
$=2 \sin \left(\frac{80^{\circ}-20^{\circ}}{2}\right) \cos \left(\frac{80^{\circ}+20^{\circ}}{2}\right) \quad\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$
$=2 \sin 30^{\circ} \cos 50^{\circ}$
$=2 \times \frac{1}{2} \cos 50^{\circ}$
$=\cos 50^{\circ}$
$=\operatorname{RH} S$
Hence, LHS = RHS.
(viii) Consider LHS :
$\sin 51^{\circ}+\cos 81^{\circ}$
$=\sin 51^{\circ}+\cos \left(90^{\circ}-9^{\circ}\right)$
$=\sin 51^{\circ}+\sin 9^{\circ}$
$=2 \sin \left(\frac{51^{\circ}+9^{\circ}}{2}\right) \cos \left(\frac{51^{\circ}-9^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$
$=2 \sin 30^{\circ} \cos 21^{\circ}$
$=2 \times \frac{1}{2} \cos \left(21^{\circ}\right)$
$=\cos \left(21^{\circ}\right)$
$=\mathrm{RHS}$
Hence, LHS = RHS.